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5(2x+3)=2x(3x+1)
We move all terms to the left:
5(2x+3)-(2x(3x+1))=0
We multiply parentheses
10x-(2x(3x+1))+15=0
We calculate terms in parentheses: -(2x(3x+1)), so:We get rid of parentheses
2x(3x+1)
We multiply parentheses
6x^2+2x
Back to the equation:
-(6x^2+2x)
-6x^2+10x-2x+15=0
We add all the numbers together, and all the variables
-6x^2+8x+15=0
a = -6; b = 8; c = +15;
Δ = b2-4ac
Δ = 82-4·(-6)·15
Δ = 424
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{106}}{2*-6}=\frac{-8-2\sqrt{106}}{-12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{106}}{2*-6}=\frac{-8+2\sqrt{106}}{-12} $
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